Integrate it! Part- III

Calculus Level 5

∫ 0 π d x ( 2 − cos ⁡ x ) 2 \int_{0}^{π} \frac{dx}{(2 - \cos x )^{2}}

Evaluate the integral above.


The answer is 1.20.

Avineil Jain by Avineil Jain , 24, Netherlands

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2 solutions

Avineil Jain
Apr 11, 2014

Consider the following integral- \textbf{Consider the following integral-}

I = ∫ 0 π d x a − c o s x I = \int_{0}^{π}\frac{dx}{a-cosx}

It is pretty easy to evaluate I \textbf{It is pretty easy to evaluate I} c o s x = 1 − t a n 2 x 2 1 + t a n 2 x 2 cosx = \frac{1-tan^{2}\frac{x}{2}}{1+ tan^{2}\frac{x}{2}}

Simplifying, \textbf{Simplifying,}

I = ∫ 0 π s e c 2 x 2 d x ( a − 1 ) + ( a + 1 ) t a n 2 x 2 I = \int_{0}^{π}\frac{sec^{2}\frac{x}{2}dx}{(a-1) + (a+1)tan^{2}\frac{x}{2}}

So, take tan x/2 = t and obtain- \textbf{So, take tan x/2 = t and obtain-}

∫ 0 π d x a − c o s x = π a 2 − 1 \int_{0}^{π}\frac{dx}{a-cosx}= \frac{π}{\sqrt{a^{2} - 1}}

Differentiating both sides with respect to "a", \textbf{Differentiating both sides with respect to "a",}

∫ 0 π d x ( a − c o s x ) 2 = π a ( a 2 − 1 ) 3 / 2 \int_{0}^{π}\frac{dx}{(a-cosx)^{2}} = \frac{πa}{(a^{2}-1)^{3/2}}

Substitute a=2, \textbf{Substitute a=2,}

∫ 0 π d x ( 2 − c o s x ) 2 = 2 π 3 3 \int_{0}^{π}\frac{dx}{(2-cosx)^{2}} = \frac{2π}{3\sqrt{3}}

29 Helpful 0 Interesting 0 Brilliant 0 Confused

but when i calculate this integral in calculator it comes out to be 3.1385........is it wrong....?

Manit Mukhopadhyay - 7 years, 1 month ago

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you have your calculator set to degrees and not radians

Luis Valdivia - 4 years, 6 months ago

Actually a great solution!

Carlos David Nexans - 6 years, 10 months ago

You should add a backslash before all functions including cos and tan in LaTex. See the difference \cos x " cos ⁡ x \cos x ". The function name cos is not in italic which is for constants and variables and there is a space between cos and x. Now cos x " c o s x cos x ", all are in italic and there is no space between cos and x. You don't need to enter text in LaTex. It is Brilliant.org's standard.

Chew-Seong Cheong - 3 years, 1 month ago

Nice solution. We can also solve it by integration by parts: ∫ 0 π d d x ( − 1 2 − cos ⁡ x ) × 1 sin ⁡ x d x \int_0^\pi \frac d{dx}\left(\frac{-1}{2-\cos x}\right)\times \frac 1{\sin x}dx

Sravanth C. - 3 years ago
Hassan Abdulla
Apr 30, 2018

I = ∫ 0 π d x ( 2 − cos ⁡ x ) 2 l e t t = tan ⁡ ( x 2 ) I=\int _{ 0 }^{ π } \frac { dx }{ (2-\cos x)^{ 2 } } \\ \\ let\quad t=\tan { \left( \frac { x }{ 2 } \right) } weierstrass substitution

cos ⁡ x = 1 − t 2 1 + t 2 d x = 2 d t 1 + t 2 I = ∫ 0 ∞ 2 d t ( 2 − 1 − t 2 1 + t 2 ) 2 ⋅ 2 d t 1 + t 2 = 2 ∫ 0 ∞ 1 + t 2 ( 2 − 1 − t 2 1 + t 2 ) 2 ( 1 + t 2 ) 2 d t I = 2 ∫ 0 ∞ 1 + t 2 ( 2 + t 2 − 1 + t 2 ) 2 d t = 2 ∫ 0 ∞ 1 + t 2 ( 1 + 3 t 2 ) 2 d t l e t t = tan ⁡ ( x ) 3 ⇒ d t = sec ⁡ 2 ( x ) 3 d x I = 2 ∫ 0 π 2 ( 1 + tan ⁡ 2 ( x ) 3 ) ( 1 + tan ⁡ 2 ( x ) ) 2 ⋅ sec ⁡ 2 ( x ) 3 d x = 2 3 3 ∫ 0 π 2 ( 3 + tan ⁡ 2 ( x ) ) sec ⁡ 2 ( x ) d x I = 2 3 3 ∫ 0 π 2 ( 3 cos ⁡ 2 ( x ) + sin ⁡ 2 ( x ) ) d x = 2 3 3 ∫ 0 π 2 ( 2 cos ⁡ 2 ( x ) + 1 ) d x I = 2 3 3 ∫ 0 π 2 ( 2 + cos ⁡ ( 2 x ) ) d x = 2 3 3 ( 2 x + sin ⁡ ( 2 x ) 2 ∣ 0 π 2 ) = 2 π 3 3 \cos x=\frac { 1-t^{ 2 } }{ 1+t^{ 2 } } \\ dx=\frac { 2dt }{ 1+t^{ 2 } } \\ I=\int _{ 0 }^{ \infty } \frac { 2dt }{ \left( 2-\frac { 1-t^{ 2 } }{ 1+t^{ 2 } } \right) ^{ 2 } } \cdot \frac { 2dt }{ 1+t^{ 2 } } =2\int _{ 0 }^{ \infty } \frac { 1+t^{ 2 } }{ \left( 2-\frac { 1-t^{ 2 } }{ 1+t^{ 2 } } \right) ^{ 2 }\left( 1+t^{ 2 } \right) ^{ 2 } } dt\\ I=2\int _{ 0 }^{ \infty } \frac { 1+t^{ 2 } }{ \left( 2+t^{ 2 }-1+t^{ 2 } \right) ^{ 2 } } dt=2\int _{ 0 }^{ \infty } \frac { 1+t^{ 2 } }{ \left( 1+3t^{ 2 } \right) ^{ 2 } } dt\\ let\quad t=\frac { \tan { \left( x \right) } }{ \sqrt { 3 } } \Rightarrow dt=\frac { \sec ^{ 2 }{ \left( x \right) } }{ \sqrt { 3 } } dx\\ I=2\int _{ 0 }^{ \frac { π }{ 2 } }{ \frac { \left( 1+\frac { \tan ^{ 2 }{ \left( x \right) } }{ 3 } \right) }{ \left( 1+\tan ^{ 2 }{ \left( x \right) } \right) ^{ 2 } } } \cdot \frac { \sec ^{ 2 }{ \left( x \right) } }{ \sqrt { 3 } } dx=\frac { 2 }{ 3\sqrt { 3 } } \int _{ 0 }^{ \frac { π }{ 2 } }{ \frac { \left( 3+\tan ^{ 2 }{ \left( x \right) } \right) }{ \sec ^{ 2 }{ \left( x \right) } } dx } \\ I=\frac { 2 }{ 3\sqrt { 3 } } \int _{ 0 }^{ \frac { π }{ 2 } }{ \left( 3\cos ^{ 2 }{ \left( x \right) } +\sin ^{ 2 }{ \left( x \right) } \right) dx } =\frac { 2 }{ 3\sqrt { 3 } } \int _{ 0 }^{ \frac { π }{ 2 } }{ \left( 2\cos ^{ 2 }{ \left( x \right) } +1 \right) dx } \\ I=\frac { 2 }{ 3\sqrt { 3 } } \int _{ 0 }^{ \frac { π }{ 2 } }{ \left( 2+\cos { \left( 2x \right) } \right) dx } =\frac { 2 }{ 3\sqrt { 3 } } \left( 2x+\frac { \sin { \left( 2x \right) } }{ 2 } |_{ 0 }^{ \frac { π }{ 2 } } \right) =\frac { 2\pi }{ 3\sqrt { 3 } }

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